Answer is D breaking apart I to not more than two
Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4
Mass of 1 mole of H2SO4= 98.1g
Therefore mass of 7.5x10^23 molecules of H2SO4= 122.63g
C12H22O11 aka carbon, hydrogen, and oxygen
Balanced equation:
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
<span>69.0 g HCl / 36.46 g/mol = 1.89 mol HCl </span>
<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>
<span>Calculate moles and mass CaCl2: </span>
<span>1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2</span>
Answer:
b) Delta S < 0
Explanation:
The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).
- If Δn(g) > 0, the entropy increases (ΔS > 0).
- If Δn(g) < 0, the entropy decreases (ΔS < 0).
- If Δn(g) = 0, there is little or no change in the entropy
Let's consider the following equation.
2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)
Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.
